Water vacuum evaporation thermodynamics

Question

A vacuum evaporation system works keeping warm water inside a sealed vessel, which is evacuated with a reversed compressor (or vacuum pump) to keep an inside pressure lower than atmospheric pressure, with the objective of lowering evaporation temperature and increase overall process efficiency. Next picture shows a highly optimized vacuum evaporator system schematic (for the subject of this question we can ignore all heat recovery subsystems):

I am interested in an ambient temperature vacuum evaporator (i.e. 25 ºC). Vapor pressure of water at this temperature is 3.17 kPa, so we need an inside pressure lower than or equal to this value.

There are two main sources of energy expenses. First is enthalpy of evaporation, which at this conditions is 2440 kJ/kg, so we need to add this amount of energy to maintain water temperature at 25 ºC during process. Second is compressor work to keep process pressure inside vacuum vessel. We can approximate it as a reversible adiabatic thermodynamic process (i.e. isentropic process).

Entropy of saturated water vapor at 25 ºC and 3.17 kPa (the one that compressor evacuates because liquid water stays in the bottom of the tank) is 8.56 kJ/kg. Internal energy at same conditions is 2410 kJ/kg. Because process is isentropic, entropy of discharge steam is the same, so its thermodynamic state is 400 ºC 101 kPa and internal energy is 2970 kJ/kg. Because process is adiabatic, compressor work is equal as change in internal energy: 2970 - 2410 = 560 kJ/kg.

Energy needed to run an idealized system as the one described will be the sum of the enthalpy of evaporation and the compressor work: 2440 + 560 = 3000 kJ/kg.

Is this estimation correct?

400 ºC compressor discharge steam looks quite hot to me. This can explain why most common industrial vacuum evaporation systems work at higher water temperatures and lower vacuum levels. It is right?

Answer 1

The temperature sounds correct, the energy balance does not. I get 405°C for the compressor discharge but the enthalpy increases from 2547 kJ/kg to 3289 kJ/kg, for a work input to the gas of 743 kJ/kg, not 560 kJ/kg.

I don't remember why you need to use enthalpy instead of internal energy in a compressor. But it is always correct and just as easy to look up, so you might as well always use enthalpy.

As for your final question, compressing low density gases at high pressure ratios typically results in very high discharge temperatures, so this result makes sense. This doesn't even account for the fact that realistically this compressor is going to be very inefficient and will require significant cooling or the outlet will be even hotter.